Simplify the following expression and state the condition under which the simplification is valid. You can assume that $k \neq 0$. $a = \dfrac{5k}{2k - 12} \times \dfrac{3(k - 6)}{k} $
When multiplying fractions, we multiply the numerators and the denominators. $a = \dfrac{ 5k \times 3(k - 6) } { (2k - 12) \times k } $ $ a = \dfrac {5k \times 3(k - 6)} {k \times 2(k - 6)} $ $ a = \dfrac{15k(k - 6)}{2k(k - 6)} $ We can cancel the $k - 6$ so long as $k - 6 \neq 0$ Therefore $k \neq 6$ $a = \dfrac{15k \cancel{(k - 6})}{2k \cancel{(k - 6)}} = \dfrac{15k}{2k} = \dfrac{15}{2} $